the parabola has a y-intercept of 5 anf contains the point (2,-3) and (-1,12). what are the coordinates of the vertex?

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the parabola has a y-intercept of 5 anf contains the point (2,-3) and (-1,12). what are the coordinates of the vertex?

Mathematics
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(0,5) (2,-3) (-1,12) right?
yes
0a + 0b + 5 = 5 4a +2b + 5 = -3 a -b +5 = 12 ; a = 12-5 +b ---------------------------- a = 7 +b ---------------- 4(7+b) +2b + 5 = -3 28 +4b +2b = -8 6b = -36 b = -6 c = 5 ----------------------- a = 12-5 +b; a = 12-5-6 a = 1 -------------------------- y =x^2 -6x +5 is your equation for hese points

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Other answers:

the vertex is at -b/2a 6/2 = 3 the vertex is at (3,y) y = (3)^2-6(3) +5 y = 9 -18 +5 y = 14 -18 y = -4 vartex is at (3,-4) if I did it right ;)
you agree or disagree :)
thanx ..that is the rite answer i agree
yay!! if you ever wanna know "how" its the right answer... I can do that too :)
ya..i dont get the first few steps
we know that these three points are not in a straight line, so that only leaves a parabola as an answer that we can actually get.
a parabola usually is of the form: ax^2 +bx + c = y ; and you normally try to figure out the "x" and "y" values...but they already give you those right?
yup
so we get the first 3 equations: a(0)^2 +b(0) +c = 5 ; (0,5).. c = 5 naturally
the "variables" we are trying to get them to have on common are the a, b, and c parts so this is almost a backwards solution lol
a(2)^2 + b(2) + c = -3 4a +2b +5 = -3 4a +2b = -8 right?
yup..
a(-1)^2 +b(-1) + c = 12 ; (-1,12) a -b +5 = 12 a - b = 12 - 5 a = 7 +b right?
oh..i get it!!
4a +2b = -8 ; so plub if our "value" for "a" 4(7+b) +2b = -8 28 +4b +2b = -8 6b = -36 b = -6 :)
a = 7+b so... a = 7 -6 = 1 our quad equation becomes: x^2 -6x +5 = y :)
thanx a lot friend..
youre welcome :)
would u like to look through the other question i posted too..
if I can find it lol
okay

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